leetcode 37. Sudoku Solver

/*
leetcode 37. Sudoku Solver
Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

题目大意：解数独问题
解题思路：回溯，对于每一个没有填充的位置，依次填入0...9，看是不是满足数独的条件。
检查函数：check函数检查board是否满足数独的条件。先检查每一行和每一列处理自己外，有没有和其他位置的值相等的值
然后检查自己所在3*3的方格内有没有和自己相等的值。如果都没有，说明这次检查是正确的，即到目前位置，该数独是成立的。
解决函数：对于每一个位置，如果不为'.'，说明不需要填充这个方格，继续填下一个。如果为'.',则依次填入0...9,此时需要
该处的check成立并且solve下一个位置也成立。

*/

#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Solution {
public:
void solveSudoku(vector<vector<char>>& board)
{
solve(board, 0);
}

bool solve(vector<vector<char>>& board, int position)
{
if (position == 81)
return true;

int row = position / 9;
int col = position % 9;

if (board[row][col] != '.')
return solve(board, position + 1);

for (int i = 1; i <= 9; ++i)
{
board[row][col] = i + '0';
if (check(board, row, col) && solve(board, position + 1))
return true;
board[row][col] = '.';
}

return false;
}

bool check(vector<vector<char>>& board, int x, int y)
{
//判断行和列是不是有相等的值
for (int i = 0; i < 9; ++i)
{
if (i != y && board[x][i] == board[x][y])
return false;
if (x != i && board[i][y] == board[x][y])
return false;
}

//判断3*3格子里面是不是有相等的值
for (int i = 0; i < 3; ++i)
{
for (int j = 0; j < 3; ++j)
{
if (board[x / 3 * 3 + i][y / 3 * 3 + j] == board[x][y] &&
(x != x / 3 * 3 + i) && (y != y / 3 * 3 + j))
return false;
}
}

return true;
}
};

void test()
{
Solution sol;
vector<vector<char>> board{
{'.','.','9','7','4','8','.','.','.'},
{'7','.','.','.','.','.','.','.','.'},
{'.','2','.','1','.','9','.','.','.'},
{'.','.','7','.','.','.','2','4','.'},
{'.','6','4','.','1','.','5','9','.'},
{'.','9','8','.','.','.','3','.','.'},
{'.','.','.','8','.','3','.','2','.'},
{'.','.','.','.','.','.','.','.','6'},
{'.','.','.','2','7','5','9','.','.'}
};

sol.solveSudoku(board);

for (auto v : board)
{
for (auto i : v)
cout << i << " ";
cout << endl;
}
}

int main()
{
test();

return 0;
}